red 字符 right include nal order mco har 長度

1031. Hello World for U (20)

時間限制 400 ms 內存限制 65536 kB 代碼長度限制 16000 B 判題程序 Standard 作者 CHEN, Yue

Given any string of N (>=5) characters, you are asked to form the characters into the shape of U. For example, "helloworld" can be printed as:

h  d
e  l
l  r
lowo

That is, the characters must be printed in the original order, starting top-down from the left vertical line with n₁ characters, then left to right along the bottom line with n₂ characters, and finally bottom-up along the vertical line with n₃ characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n₁ = n₃ = max { k| k <= n₂ for all 3 <= n₂ <= N } with n₁ + n₂ + n₃ - 2 = N.

Input Specification:

Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.

Output Specification:

For each test case, print the input string in the shape of U as specified in the description.

Sample Input:

helloworld!

Sample Output:

h   !
e   d
l   l
lowor

思路

對稱U型格式輸出一個字符串。

窮舉找出滿足 n1 + n3 + n2 = N + 2(3 <= n2 <= N , n3 = n1 <= n2) 的n1,n3最大值即可。

代碼

#include<iostream>
#include<string>
#include<math.h>
using namespace std;
int main()
{
  string s;
  while(cin >> s)
  {
      int N = s.size();
      int maxn3 = -1;
      for(int n2 = 3;n2 <= N;n2++)
      {
          for(int n3 = 0;n3 <=n2;n3++)
          {
              if(2*n3 + n2 - 2 == N)
                maxn3 = max(maxn3,n3);
          }
      }

      int n1 = maxn3,n2 = N + 2 - 2 * n1;

      //print
      int i = 0;
      for(i = 0;i < n1 - 1;i++)
      {
          cout << s[i];
          for(int j = 0; j < n2 - 2;j++)
            cout << " ";
          cout << s[N - i - 1] << endl;
      }
      int rest = N - i - 1;
      for(;i <= rest;i++)
      {
          cout << s[i];
      }
      cout << endl;
  }
}

Pat1031:Hello World for U